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| /* | |
| 3n+1 問題 | |
| */ | |
| using namespace std; | |
| int dp[1000001] = {1, 1}; | |
| int calculate(int n) { | |
| // 大於 10^6 的就不暫存直接暴力算 | |
| // 因為會很快就收斂到較小的數值 | |
| if (n > 1000000) { | |
| if (n & 1) | |
| return calculate(3 * n + 1) + 1; | |
| else | |
| return calculate(n / 2) + 1; | |
| } | |
| // 使用動態規劃避免重複計算 | |
| if (dp[n]) | |
| return dp[n]; | |
| if (n & 1) | |
| dp[n] = calculate(3 * n + 1) + 1; | |
| else | |
| dp[n] = calculate(n / 2) + 1; | |
| return dp[n]; | |
| } | |
| int main() { | |
| cin.tie(0); | |
| ios_base::sync_with_stdio(0); | |
| int i, j; | |
| while (cin >> i >> j) { | |
| cout << i << " " << j << " "; | |
| // 這題 i 不一定小於 j | |
| // 這點要注意一下 | |
| if (i > j) | |
| swap(i, j); | |
| int ans = 0; | |
| for (int k = i; k <= j; k++) | |
| ans = max(ans, calculate(k)); | |
| cout << ans << "\n"; | |
| } | |
| return 0; | |
| } |